Solenoid is proving a little bit confusing

While getting through solenoid I found that the field outside it is extremely small and is negligible. Also the field at ends is half of that of center. When I google it I doesn't found any satisfying explanation that why the field at ends is half that of center and almost zero outside.

A solenoid I have got the following explanation and thoughts of mine on my question

  • Since current is flowing in upward direction on one side and downward in other hence they might cancel each other. But i have got another doubt here that since the solenoid can be thought of as many circular rings and coil has not zero field outside it.So how it can be possible in solenoid.

  • I didn't get the explanation that why at edges the field is half. Please correct me if I am wrong.

Thanks in advance!

  • $\begingroup$ What makes you think the field is 1/2 at the ends? $\endgroup$
    – Bob D
    Commented Jan 11, 2019 at 15:15
  • $\begingroup$ It is given in my book. $\endgroup$
    – David Wax
    Commented Jan 11, 2019 at 15:18
  • $\begingroup$ It is indeed half of what it is at centre of the solenoid but only if we consider a solenoid of finite length. $\endgroup$ Commented Jan 11, 2019 at 15:32
  • $\begingroup$ @AdityaGarg You mean infinite length don't you? $\endgroup$ Commented Dec 1, 2019 at 13:53

2 Answers 2


It's easy to see why the flux density (field strength) at the end is half that in the middle, for a long solenoid (length >> diameter). [Edit prompted by comment below: this, and what follows, applies only to axial field component.]

For a given number of turns per unit length, the field in the middle (or many diameters away from the ends) is independent of the length of the solenoid. This is because, if the solenoid were extended by adding extensions to either end, these extensions would be too far from the middle to affect the field there!

Now, the field in the middle of the solenoid is the sum of fields due to the left hand half and the right hand half. But, by symmetry, the fields due to these two halves are equal. So the fields strengths at the ends of these two halves are equal to half the field strength in the middle of the original solenoid.

You might object that the field at the left hand end of the long solenoid isn't just the field due to the left hand half of the long solenoid. But it is! The right hand half (and, for that matter, all but the left hand part of the left hand half!) is too far away to affect the field at the left hand end. 'Long' means … long!

  • 2
    $\begingroup$ I think this reasoning applies only to the component of the field along the axis of the solenoid. At the edge of the semi infinite solenoid, there is also a radial component that cannot be obtained by symmetry arguments. $\endgroup$ Commented Jan 11, 2019 at 16:19
  • 1
    $\begingroup$ You're quite right, of course. At the ends of the solenoid, if we go off-axis, the field has a radial component. In the middle of the solenoid the field is parallel to the axis so the radial component, even off axis, is zero. If we wanted to, we could see this as due to the radial fields due two halves of the solenoid cancelling! $\endgroup$ Commented Jan 11, 2019 at 16:41
  • $\begingroup$ What about my first part $\endgroup$
    – David Wax
    Commented Jan 13, 2019 at 9:38

The point about "half the value" at the end-point has been answered.

About your first point: the discussion is about a very long solenoid, as explained above.

Indeed, the field that comes out of a very long solenoid will come back to reenter through the other end; os it is nor strictly zero outside, it does hava a component along the axis of the solenoid, in the opposite direction. But the lines of field will be very long, and make very wide curves. For a very long solenoid, the field will be so diluted as to be negligible.


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